$h(t) = -7t+2(f(t))$ $f(n) = 2n^{2}+3n-7$ $ h(f(1)) = {?} $
First, let's solve for the value of the inner function, $f(1)$ . Then we'll know what to plug into the outer function. $f(1) = 2(1^{2})+(3)(1)-7$ $f(1) = -2$ Now we know that $f(1) = -2$ . Let's solve for $h(f(1))$ , which is $h(-2)$ $h(-2) = (-7)(-2)+2(f(-2))$ To solve for the value of $h$ , we need to solve for the value of $f(-2)$ $f(-2) = 2(-2)^{2}+(3)(-2)-7$ $f(-2) = -5$ That means $h(-2) = (-7)(-2)+(2)(-5)$ $h(-2) = 4$